Please tell me what I'm doing wrong...
The side lengths of tops of three squares tables can be described as three consecutive integers. The combined area of the table tops is 677 sq in.
Algebraically, determine the roots of the quadratic equation. (I'll use quadratic, but if it's easier factoring or completing the square, please tell me!)
3x^2 - 6x -677 = 0
6 +- sqrt 36- 4(3)(-677)
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6
x = 6 +-sqrt 510
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6
(In this step, the sixes can be removed by division, so I don't have the proper roots)
Any help is really appreciated!
If the middle table has side x, then the three squares
(x-1)^2 + x^2 + (x+1)^2 = 677
3x^2 + 2 = 677
3x^2 = 675
x^2 = 225
x = 15
and the tables have sides 14,15,16
how did you come up with 3x^2-6x = 677?
To find the roots of the quadratic equation 3x^2 - 6x - 677 = 0, you can use the quadratic formula. The formula is:
x = (-b ± √(b^2 - 4ac)) / (2a)
In the given equation, a = 3, b = -6, and c = -677. Now, let's substitute these values into the quadratic formula:
x = (-(-6) ± √((-6)^2 - 4(3)(-677))) / (2(3))
Simplifying further:
x = (6 ± √(36 + 8136)) / 6
x = (6 ± √8172) / 6
Now, you can simplify the square root:
x = (6 ± √(36 * 227)) / 6
x = (6 ± 6√227) / 6
Finally, you can simplify by canceling out the common factor of 6:
x = 1 ± √227
So, the roots of the quadratic equation 3x^2 - 6x - 677 = 0 are:
x = 1 + √227 and x = 1 - √227